The numbers from 1 to 100 are arranged, snaking back and forth from the bottom, in a 10x10 grid:

100 99 98 97 96 95 94 93 92 91

 81 82 83 84 85 86 87 88 89 90

 80 79 78 77 76 75 74 73 72 71

 61 62 63 64 65 66 67 68 69 70

 60 59 58 57 56 55 54 53 52 51
+--------------+
|41 42 43 44 45|46 47 48 49 50
|              |
|40 39 38 37 36|35 34 33 32 31
|              |         
|21 22 23 24 25|26 27 28 29 30
|              |         
|20 19 18 17 16|15 14 13 12 11
+--+           |
| 1| 2  3  4  5| 6  7  8  9 10
+--+-----------+

Marked off are two subsquares that, if you add up the numbers within them, the total is the square of one of the numbers within. The larger square (5x5) totals 625, the square of 25, which is found within that marked square. The other one shown is the trivial "1", a 1x1 square, the single number in which of course is its own square.

Find another subsquare where the sum of its contained numbers is equal to the square of one of those numbers.

I made the following:

Calling the columns "c", from left to right the numbers in the rows are of the form (from bottom, upwards), in order:

c, (21-c), (20+c), (41-c), (40+c), (61-c), (60+c), (81-c), (80+c), and (101-c).

When we sum the numbers of an "even" matrix (2x2, 4x4, 6x6, 8x8, 10x10, the sum doens´t depend on c, and the sums achieved are never perfect square:

2x2 .... sums 42/82/122/162/202/242/282/322/362

4x4..... sums 328/428/728/808/968/1128/1288

6x6 .... sums 1098/1458/1818/2178/2538

8x8 ..... sums 2592/3232/3872

10x10.. sum 5050

--------------------------------------------------------------

The sums for an "odd" matrix, depends on c, and are:

3x3 ...... (786-3c)/(663+3c)/(606-3c)/(483+3c)/(426-3c)/(303+3c)/(246-3c)/(123+3c)

5x5...... (1915-5c)/(1610+5c)/(1415-5c)/(1110+5c)/(1015-5c)/(610+5c

7x7..... (3248-7c)/(2681+7c)/(2268-7c)/(1701+7c)

9x9 .... (4545-9c)/(3636+9c)

------------------------------------------------

Varying c from 1 to 10, the squares found was 606-3*10 =576, 303+3*7=324, 246-3*7 = 225, e 123 +3*7 = 144. The only one that fits the rules is the underlined.

----------------------------------------------- 

Maybe, an interesting version of this problem should be a great matrix m x m...

Edited on May 23, 2008, 5:20 pm

Edited on May 23, 2008, 5:45 pm

Posted by pcbouhid on 2008-05-23 14:28:04